Modern Physics 7 Question 46
51. A nucleus at rest undergoes a decay emitting an $\alpha$-particle of de-Broglie wavelength, $\lambda=5.76 \times 10^{-15} m$. If the mass of the daughter nucleus is $223.610 amu$ and that of the $\alpha$-particle is $4.002 amu$. Determine the total kinetic energy in the final state. Hence obtain the mass of the parent nucleus in amu.
$\left(1 amu=931.470 MeV / c^{2}\right)$
$(2001,5 M)$
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Answer:
Correct Answer: 51. $6.25 \times 10^{11}$, zero, $5.0 eV$
Solution:
- (a) Given mass of $\alpha$-particle, $m=4.002$ amu and mass of daughter nucleus,
$$ M=223.610 amu $$
de-Broglie wavelength of $\alpha$-particle,
$$ \lambda=5.76 \times 10^{-15} m $$
So, momentum of $\alpha$-particle would be
$$ \begin{aligned} p & =\frac{h}{\lambda}=\frac{6.63 \times 10^{-34}}{5.76 \times 10^{-15}} kg-m / s \\ \text { or } \quad p & =1.151 \times 10^{-19} kg-m / s \end{aligned} $$
From law of conservation of linear momentum, this should also be equal to the linear momentum of the daughter nucleus (in opposite direction).
Let $K _1$ and $K _2$ be the kinetic energies of $\alpha$-particle and daughter nucleus. Then total kinetic energy in the final state is
$$ \begin{aligned} K & =K _1+K _2=\frac{p^{2}}{2 m}+\frac{p^{2}}{2 M}=\frac{p^{2}}{2} \frac{1}{m}+\frac{1}{M} \\ K & =\frac{p^{2}}{2} \frac{M+m}{M m} \\ 1 amu & =1.67 \times 10^{-27} kg \end{aligned} $$
Substituting the values, we get
or
$$ \begin{aligned} & K=10^{-12} J \\ & K=\frac{10^{-12}}{1.6 \times 10^{-13}}=6.25 MeV \end{aligned} $$
(b) Mass defect, $\Delta m=\frac{6.25}{931.470}=0.0067 amu$
Therefore, mass of parent nucleus $=$ mass of $\alpha$-particle + mass of daughter nucleus + mass $\operatorname{defect}(\Delta m)$
$$ \begin{aligned} & =(4.002+223.610+0.0067) amu \\ & =227.62 amu \end{aligned} $$
Hence, mass of parent nucleus is $227.62 amu$.
